Two particles move on a smooth horizontal surface.
The positions, in Cartesian coordinates, of the
particles at time \(t\)
are \((a+ut\cos\alpha \,,\, ut\sin\alpha)\) and
\((vt\cos\beta\,,\, b+vt\sin\beta )\), where \(a\), \(b\), \(u\) and \(v\) are positive
constants, \(\alpha\) and \(\beta\) are constant acute angles, and \(t\ge0\).
Given that the two particles collide, show that
\[
u \sin(\theta+\alpha) = v\sin(\theta +\beta)\,,
\]
where \(\theta \) is the acute angle satisfying \(\tan\theta = \dfrac b a\).
A gun is placed on the top of a vertical tower of height \(b\) which stands on horizontal ground. The gun fires a bullet with speed \(v\) and (acute) angle of elevation \(\beta\). Simultaneously, a target is projected from a point on the ground a horizontal distance \(a\) from the foot of the tower. The target is projected with speed \(u\) and (acute) angle of elevation \(\alpha\), in a direction directly away from the tower.
Given that the target is hit before it reaches the ground, show that
\[
2u\sin\alpha (u\sin\alpha - v\sin\beta) > bg\,.
\]
Explain, with reference to part (i), why the target can only be hit if \(\alpha > \beta\).
Solution:
The particles collide if there exists a time when
\begin{align*}
&& a + ut \cos \alpha &= vt \cos \beta \\
\Rightarrow && t (v \cos \beta-u \cos \alpha) &= a\\
&& ut \sin \alpha &= b + vt \sin \beta \\
\Rightarrow && t(u \sin \alpha - v \sin \beta) &= b\\
\Rightarrow && a(u\sin \alpha - v \sin \beta) &= b(v \cos \beta - u \cos \alpha) \\
\Rightarrow && u(a \sin \alpha + b \cos \alpha) &= v (b \cos \beta + a \sin \beta) \\
\Rightarrow && u \sin (\alpha + \theta) &= v \sin (\beta + \theta)
\end{align*}
The path of the bullet is \((vt \cos \beta, b + vt \sin \beta -\frac12 g t^2)\). The path of the target is \((a+ut \cos \alpha, ut \sin \alpha - \frac12 g t^2)\).
By comparing components as in part (i) and noting the acceleration doesn't change the story, we can see that \(t(u \sin \alpha - v \sin \beta) = b\) and we also need \(u t \sin \alpha - \frac12 gt^2 >0\) or \(u \sin \alpha - \frac12 gt > 0\)
\begin{align*}
&& u \sin \alpha & > \frac12 gt \\
&& 2u \sin \alpha & > g \frac{b}{(u \sin \alpha - v \sin \beta)} \\
\Rightarrow && 2u \sin \alpha( u \sin \alpha - v \sin \beta) & > gb
\end{align*}
Notice we must have \(u \sin \alpha > v \sin \beta\) and \(u \sin (\alpha + \theta) = v \sin (\beta + \theta)\) so \( \frac{\sin \alpha}{\sin (\alpha + \theta)} > \frac{\sin \beta}{\sin (\beta + \theta)}\), but if we consider \(f(t) = \frac{\sin t}{\sin(t+x)}\) we can see \(f'(t) = \frac{\cos t \sin(t + x) - \sin t \cos(t+x)}{\sin^2(t+x)} = \frac{\sin x}{\sin^2(t+x)} > 0\) is increasing, therefore \(\alpha > \beta\).